The rounds are open and rated for everybody.
Mentioned in the second line ...

it will be a meme)

Codeforces: 20 T-shirts randomly distributed.

Me: Pretending to calculate the probability when I know answer is 0.

But scientific studies also show that its_Atrap

The statements are concise, sold it

PurpleCrayon orz

I hope to become Purple after this round :)

I vote only after contest.

Very original meme

Woaahhh, Directly trying to defeat red coders. Nice MEME!!!

OK I guess global rounds are really hard this was my first codeforces contest I guess i shouldnt be demotivated because i didnt even submit a single thing right just got close on phoenix and puzzle(failed test 2 lol)

omg :O

And a very good opportunity to boost it down, too =)

lol no. for the same performance you get much more rating in div2 than in global/div1. Not that it should matter but that's not the point here.

I have observed there's a lot more cheating in global rounds than normal rounds. Honest participants usually suffer in global rounds imo.

as long as you include "As a tester" in your msg, its fine :)

I like the strategy which has a possibility that we don't solve anything in the contest XD.

by the time I solve it, it would be worth a negative amount of points.

You guys are solving 5000 pts problem :(

All the best bro.

They're both americans with a chinese name, i guess

Daily reminder that Google exists

Same here brother, me too suffering from covid, I can feel your pain. T_T :(

Same situation. Tested positive yesterday. Unlike others I am having a headache, so decided to take some rest.

If you are not well, I advise you to do the same.

Hmmm, I was imagining a closet full of segment trees, linked lists and a few queues. Never thought of the struct t_shirt. Hmmm..

Maybe you have to do the same thing after 2 hrs in this contest(considering the difficulty of global round).

I worked like two hours on d but didnt get it :/

And it can pass the pretest.

Note that there cannot be more than one consecutive light that turns on automatically. So, our final answer will have some segments of $$$1$$$'s (lights that are turned on manually) separated by exactly one $$$0$$$ (lights that are turned on automatically). We need to find the number of ways to make such an arrangement.

Let's take this one step at a time.

In how many ways can we turn all the lights on manually?

You can only choose one starting light in a segment of n lights, then at each step you can either go left or go right. For better visualisation, consider n points coloured red. Pick a starting point and colour it blue. In each step, you can move from a blue point to an adjacent red point.

Number of ways to do this = $$$(L + R)! / (L! * R!)$$$ , where L is the number of steps to the left of the starting point that one can take, and similarly, R is to the right.

To meet the constraints, we will be precomputing these answers till $$$n$$$.

Now, coming to the main solution.

Let $$$dp[i][j]$$$ = number of ways we can arrange the first $$$i$$$ lights such that the $$$i$$$ $$$th$$$ light is manually turned on and a total of $$$j$$$ lights were turned on automatically among the first $$$i$$$ lights.

Let $$$k...i$$$ be the last segment of $$$1$$$'s. This means that at position $$$k-1$$$ we will have a $$$0$$$ and at position $$$k-2$$$ there will be a $$$1$$$. If there are exactly $$$j$$$ $$$zeros$$$ in the segment $$$1...i$$$ then there are exactly $$$j - 1$$$ $$$0$$$s in the segment $$$1...k-2$$$.

Let's define an operation, as we turning a light manually on (or placing a $$$1$$$). So in the segment $$$1...k-2$$$ we have to place $$$k - 2 - (j - 1)$$$ ones, and in the segment $$$k...i$$$ we have to place all $$$i - k + 1$$$ ones. This is similar to the question we answered above for placing all ones. We have to tell the number of permutations of $$$(L + R)$$$ items, in which the order of the $$$L$$$ items and $$$R$$$ items are fixed.

Therefore, the ans for $$$dp[i][j] = dp[k - 1][j - 1] * nCr((k - 2 - (j - 1)) + (i - k + 1), (i - k + 1)) * ways$$$ $$$to$$$ $$$arrange$$$ $$$(i - k + 1)$$$ $$$ones$$$

PS: I know it's a little too verbose. But hope you can understand the solution.

You can simply let computer $$$i$$$ in $$$dp[i][j]$$$ be turned on automatically.

so you just have to enumerate the last automatically turned on computer before computer $$$i$$$ for $$$dp[i][j]$$$

Video Explanation for Problem D

For b check the if number is divisible of 2 or 4 and the remaining number should be a square number.

In C, keep assigning the block to the tower with minimum current height. Can be done using set <pair<int, int > >. Also the answer can never be NO,

For C, just sort by block lengths and assign to each stack.

For B, divide by 2 and 4, and see if number is a square.

B — If I don't FST, if dividing by 2 or 4 gives you a square (number) its a yes C — no idea I kept WAing

Assuming that I will not fail system tests.

Problem B: Notice that we can only form squares using 2 or 4 isosceles triangles. So just check if the given number of triangles is divisible by 2. If it is, then check if (n/2) is a perfect square. (We can form a square of size sqrt(n) considering 2 triangles as a single block). Now, if this fails then check if n is divisible by 4. If it is, then repeat the process above. The answer in any other case is NO.

Problem C: I have done this greedily. Using a priority queue, I always select the tower with minimum height and add the blocks in decreasing order after sorting. After adding all the blocks, I check if the difference between the minimum and maximum height is smaller than or equal to x.